已知 $f(x)$ 连续，$f(a)=f(b)=0$，求证 $\exists \epsilon \in(a,b)$ s.t. $\frac{f’(\epsilon)}{-2\epsilon}=f(\epsilon)$.

证：把 $\frac{f’(\epsilon)}{-2\epsilon}=f(\epsilon)$ 看成微分方程 $\frac{y’}{-2x}=y$.

$$ \begin{align} \frac{dy}{dx\cdot(-2x)} & =y \ \frac{dy}{y} & =-2x,dx \ \ln y & =-x^{2}+C \ y & =e^{-x^{2}+C} \ y\cdot e^{x^{2}}-C’ & =0 \end{align} $$

则可以构造函数 $F(x)=f(x)\cdot e^{x^{2}}$.

由题：$F(a)=F(b)=0$

所以由罗尔定理：$\exists\epsilon \in(a,b)$ s.t. $F’(\epsilon)=f’(\epsilon)\cdot e^{\epsilon^{2}}+f(\epsilon)\cdot e^{\epsilon^{2}}\cdot 2\epsilon=0$, i.e. $\frac{f’(\epsilon)}{-2\epsilon}=f(\epsilon)$